141. Linked List Cycle

Summary

Solution Approach	Explanation (1-minute)	Time Complexity	Space Complexity
Hash Set	Store visited nodes in a set. If we encounter a node already in the set, there's a cycle.	\(O(n)\)	\(O(n)\)
Two Pointers (Floyd's Cycle Detection)	Use slow and fast pointers. Slow moves 1 step, fast moves 2 steps. If they meet, there's a cycle.	\(O(n)\)	\(O(1)\)

Implementation

PythonJavaScriptGo

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        return False

```